13.8 Summary

• The ideal gas equation connecting pressure \((P)\), volume \((V)\) and absolute temperature \((T)\) is
  \(
  P V=\mu R T=k_B N T
  \)
  where \(\mu\) is the number of moles and \(N\) is the number of molecules. \(R\) and \(k_B\) are universal constants.
  \(
  R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \quad k_B=\frac{R}{N_A}=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}
  \)
  Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures.
• The kinetic theory of an ideal gas gives the relation
  \(
  P=\frac{1}{3} n m \overline{v^2}
  \)
  where \(n\) is number density of molecules, \(m\) the mass of the molecule and \(\overline{v^2}\) is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature.
  \(
  \frac{1}{2} m \overline{v^2}=\frac{3}{2} k_B T, \quad v_{\text {rrs }}=\left(\overline{v^2}\right)^{1 / 2}=\sqrt{\frac{3 k_B T}{m}}
  \)
  This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed.
• The translational kinetic energy
  \(
  E=\frac{3}{2} k_B N T
  \)
  This leads to a relation
  \(
  P V=\frac{2}{3} E
  \)
• The law of equipartition of energy states that if a system is in equilibrium at an absolute temperature \(T\), the total energy is distributed equally in different energy modes of
  absorption, the energy in each mode being equal to \(1 / 2 k_B T\). Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy \(1 / 2 k_B\) T. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to \(2 \times 1 / 2 k_B T=k_B T\).
• Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion.
• The mean free path lis the average distance covered by a molecule between two successive collisions :
  \(
  l=\frac{1}{\sqrt{2} n \pi d^2}
  \)
  where \(n\) is the number density and d the diameter of the molecule.
• The pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer.
• We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule.
• The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is \(1 / 2 k_B\) T. Each quadratic term in the total energy expression of a molecule is to be counted as a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy \(2 \times 1 / 2 k_B T=k_B T\)
• Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy (mgh) for ordinary heights is much less than the average kinetic energy \(1 / 2 m v^2\) of the molecules. \(\left\langle v^2\right\rangle\) is not always equal to \((\langle v\rangle)^2\). The average of a squared quantity is not necessarily the square of the average. 

Question For Short Answer

Q1. When we place a gas cylinder on a van and the van moves, does the kinetic energy of the molecules increase? Does the temperature increase?

Solution:

No, the kinetic energy of the molecules does not increase. This is because the velocity of the molecules does not increase with respect to the walls of the gas cylinder when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas’s kinetic energy and there will be a rise in the temperature.

Q2. While gas from a cooking gas cylinder is used, the pressure does not fall appreciably till the last few minutes. Why?

Solution:

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. The boiling point of a liquid depends on the pressure above its surface. The higher the pressure above the liquid, the higher will be its boiling point.
When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate for this fall in pressure, more liquid undergoes a phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.

Q3. Do you expect the gas in a cooking gas cylinder to obey the ideal gas equation?

Solution:

No, the gas won’t obey the ideal gas equation due to the following reasons:
1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder because the liquid state does not obey the ideal gas equation.

Q4. Can we define the temperature of (a) vacuum, (b) a single molecule?

Solution:

(a) Temperature is defined as the average kinetic energy of the particles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities.
(b) No, we cannot define the temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules.

Q5. Comment on the following statement: the temperature of all the molecules in a sample of a gas is the same.

Solution:

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because the temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, the temperature will be the same for all the molecules.

Q6. Consider a gas of neutrons. Do you expect it to behave much better as an ideal gas as compared to hydrogen gas at the same pressure and temperature?

Solution:

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:
1. As per the Kinetic theory, neutrons do not interact with each other. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.
2. Neutrons are smaller than hydrogen. This fulfills another kinetic theory postulate that gas molecules should be points and should have negligible size.

Q7. A gas is kept in a rigid cubical container. If a load of 10 kg is put on top of the container, does the pressure increase?

Solution:

No, the pressure on gas won’t increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.

Q8. If it were possible for a gas in a container to reach the temperature 0 K, its pressure would be zero. Would the molecules not collide with the walls? Would they not transfer momentum to the walls?

Solution:

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because the pressure of a gas is formed due to the molecule’s collision with the walls of the container.

Q9. It is said that the assumptions of the kinetic theory are good for gases having low densities. Suppose a container is so evacuated that only one molecule is left in it. Which of the assumptions of kinetic theory will not be valid for such a situation? Can we assign a temperature to this gas?

Solution:

Two postulates of kinetic theory will not be valid in this case. These are given below:
1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction, and time.

Q10. A gas is kept in an enclosure. The pressure of the gas is reduced by pumping out some gas. Will the temperature of the gas decrease by Charles’s law?

Solution:

If the gas is ideal, there will be no temperature change. Moreover, Charles’s law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles’s law.

Q11. Explain why cooking is faster in a pressure cooker.

Solution:

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means the boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at \(100^{\circ} \mathrm{C}\). Thus, the cooking process gets faster.

Q12. If the molecules were not allowed to collide among themselves, would you expect more evaporation or less evaporation?

Solution:

If the molecules are not allowed to collide with each other, they will have long mean free paths, and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at a very low temperature.

Q13. Is it possible to boil water at room temperature, say \(30^{\circ} \mathrm{C}\)? If we touch a flask containing water boiling at this temperature, will it be hot?

Solution:

Yes, it is possible to boil water at \(30^{\circ} \mathrm{C}\) by reducing the external pressure. A liquid boils when its vapour pressure equals the external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.
No, the flask containing water boiling at \(30^{\circ} \mathrm{C}\) will not be hot.

Q14. When you come out of a river after a dip, you feel cold. Explain.

Solution:

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body. As a result of it, the temperature of our body falls down due to the loss of heat and we feel cold.

Q15. Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through windows, doors, etc.

Solution:

We Know internal energy at a particular temperature
\(
U=n C_v T
\)
Air in the home is are chiefly diatomic molecules, so
\(
\begin{aligned}
& C_v=\frac{5}{2} R \\
& \therefore U=\frac{5 n}{2} R T
\end{aligned}
\)
Now by eqn. of state
\(
\begin{aligned}
& n R T=P V \\
& U=\frac{5}{2}(n R T) \\
& \Rightarrow U=\frac{5}{2} P V
\end{aligned}
\)
Now pressure \(\mathrm{P}\) is Constant also \(\mathrm{V}\) of the room = constant
Thus,
\(U=\) Constant